Titration answer 6

Lack of iron in the diet is a common cause of anaemia. This can be corrected by taking "iron" tablets, which contain ferrous sulfate, now known as iron(II) sulfate. In this exercise, we analyze the iron content of two such ferrous sulfate tablets.

Label from a bottle of iron tablets

Two ferrous sulfate tablets were ground to a paste with 1 M sulfuric acid using a mortar and pestle. The paste was made up to 100 cm³ in a volumetric flask with further 1 M sulfuric acid. After thorough mixing, 10 cm³ samples of this solution were titrated with 0.00200 M potassium manganate(VII) solution. The average titre was 23.30 cm³.

Why is such a large volume of 1 M sulfuric acid used in this experiment?

Calculate the mass of anhydrous iron(II) sulfate in one tablet.

Why might your answer disagree with the value quoted on the bottle?

Brown colouration due to incomplete reduction of the manganate(VII) ion

The sulfuric acid is needed to provide the 8H⁺ shown in the equation. Failure to add sufficient acid leads to incomplete reduction of the manganate(VII) ion. Rather than being reduced to the virtually colourless manganese(II) ion, it is often reduced to the brown solid, manganese(IV) oxide. This is not a quantitative reduction, and will prevent the calculation from being accurate.

Number of moles of manganate(VII) = 23.30/1000 × 0.00200

MnO₄^-(aq) + 8H⁺(aq) + 5Fe²⁺(aq) ⇒Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

5Fe²⁺ º MnO₄^-

\ Number of moles of iron(II) in 10cm³ = 5 × 23.30/1000 × 0.00200

\ Number of moles of iron(II) in 100cm³ = 10 × 5 × 23.30/1000 × 0.00200

FeSO₄ = 152

\ Mass of iron(II) sulfate in 100cm³ (2 tablets) = 152 × 10 × 5 × 23.30/1000 × 0.0020

\ Mass of iron(II) sulfate in 1 tablet = ½ × 152 × 10 × 5 × 23.30/1000 × 0.0020 = 0.177 g

This is 177 mg which compares to the manufacturer's stated value of 200 mg.

The smaller mass may be a result of poor technique, with some iron(II) being lost during the transfer from the mortar to the volumetric flask. Poor grinding will result in the failure to extract all the iron(II). It is also possible that some of the iron(II) has become oxidized to iron(III) over time.